转换为循环以应用 [英] Convert for loop to apply

问题描述

在R中，如何使用apply，lapply，rapply，do.call等功能替换以下代码?

In R, how do you replace the following code using functions like apply, lapply, rapply, do.call, etc.?

u <- 10:12
slist <- list()

for (i in 1:length(u)) {
  p <- combn(u, i) 
  for (j in 1:ncol(p)) {
    s <- paste(p[,j], collapse=",")
    slist[[s]] <- 0
  }
}

对于此部分:

  for (j in 1:ncol(p)) {
    s <- paste(p[,j], collapse=",")

我尝试过类似的事情:

  s <- apply(p, 2, function(x) paste(x, collapse=","))

哪个工作.但是对于同一个for循环中的slist[[s]] <- 0部分，我不知道该怎么办.

Which works. But then for that slist[[s]] <- 0 part inside that same for-loop, I don't know what to do.

编辑:这是我要尝试的操作.对于向量u，我正在生成该向量中所有子集的列表.然后，对于每个子集，我将其分配给s，然后使用字符串s作为slist中元素的名称.我知道有点奇怪，但这是用于家庭作业.对于上面的代码，这将是slist的前5个元素:

This is what I'm trying to do. For the vector u, I'm producing a list of all the subsets in that vector. Then for each subset, I'm assigning it to s, then using the string s as the name of an element in slist. Kind of strange, I know, but it's for a homework assignment. For the code above, this would be the first 5 elements of slist:

 > slist
 $`10`
 [1] 0

 $`11`
 [1] 0

 $`12`
 [1] 0

 $`10,11`
 [1] 0

 $`10,12`
 [1] 0

是的，我只是想学习如何正确使用Apply和东西.

Yeah, I'm just trying to learn how to use apply and stuff properly.

推荐答案

这里是一种解决方案:

n <- unlist(lapply(seq_along(u), function(i) {
  apply(combn(length(u),i),2, function(x) paste(u[x], collapse=','))
}
))

slist <- list()
slist[n] <- 0

更新与@djhurio同时发布，它非常相似，但是我自由更改了combn的用法，因此它可以处理长度为1的u，如@ djhurio指出.

UPDATE Posted at the same time as @djhurio, it is very similar, but I took the liberty of changing the use of combn so it handles u of length 1, as @djhurio pointed out.

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